Appendix II. 879
at E and D will respectively shew weights of dry . steam and
entrained water.
P. 605. Problems on Energy Changes in a "Working
Gas.—When a gas is expanded or compressed in any manner
behind a working piston,
Heat supplied = change of internal energy + external work
and any of the three terms may be plus, minus, or zero. Thus,
when expanding isothermally, the internal change is zero, and the
gas must be given a positive heat equal to the total work done
(p. 605); and in isothermal compression the heat supply is -
negative, that is, must be abstracted to the same amount as
before.
Example 64. Assuming the p v curve to be hyperbolic, let steam
enter the cylinder at 120 Ibs. absolute pressure per square in., tem-
perature 348° F., and expand to 30 Ibs. absolute, temperature 250° F.
Find heat supplied per Ib. weight. (Hons. Steam Exam. 1894.)
Spec. vol. at 250° = 13*5 cub. ft. and P = (144x30) Ibs.
.-. p V = 144x30 x 13*5 = 58320 ft. Ibs. ; which is constant.
120
= 4 and loger =. 1*3863
P 30
Total heat at first, Hj = S^Lj = 348-32 + 871 = 1187 B.T.U.
Total heat at end, H2 = S2H-L2 = 250-32 + 940 =1158 B,T.U.
Loss of internal energy = 772 (Hj, - P V) - 772 (H2 - P V)
= 22388 ft. Ibs.
.". Heat supplied = external work - internal change
= PVloge7-~ 22388 = (58320X1*3863)-22388
= 58461 ft. Ibs., or 747 B.T.U. per 13^ cub. it, or per I Ib. weight.
[N.B.—The real result should be 87 B.T.U., the discrepancy being
due to assuming the expansion of dry steam as hyperbolic.]
Example 65.—Find expressions for the thermal efficiency, when
atmospheric air at 60° Fahr. is compressed adiabatically, then cooled
under constant pressure to 60° Fahr., and afterwards exhausted adia-
batically while doing work, till it reaches the atmospheric pressure
again.
Referring to FHg. 840, OD and o F are scales of absolute pressure
and volume respectively. Air is drawn in along A B, compressed to c,